Find the half-angular width of the central ma | vedclass

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(B) The condition for the first minimum in a single-slit Fraunhofer diffraction pattern is given by $a \sin 	heta = n \lambda$. For the first minimum

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$30$

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(B) The condition for the first minimum in a single-slit Fraunhofer diffraction pattern is given by $a \sin 	heta = n \lambda$. For the first minimum,$n = 1$,so $\sin 	heta = \frac{\lambda}{a}$.Here,the slit width $a = 12 	imes 10^{-5} 	ext{ cm} = 12 	imes 10^{-7} 	ext{ m}$.The wavelength $\lambda = 6000 \, \mathring{A} = 6000 	imes 10^{-10} 	ext{ m} = 6 	imes 10^{-7} 	ext{ m}$.Substituting the values: $\sin 	heta = \frac{6 	imes 10^{-7}}{12 	imes 10^{-7}} = \frac{6}{12} = 0.5$.Since $\sin 	heta = 0.5$,the half-angular width $	heta = \arcsin(0.5) = 30^{\circ}$.

Find the half-angular width of the central maximum in the Fraunhofer diffraction pattern produced by a slit of width $12 \times 10^{-5} \text{ cm}$,when illuminated by monochromatic light of wavelength $6000 \, \mathring{A}$. (Answer in degrees)

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